1. **State the problem:** We are given two functions: $$f(x) = 2x + 3$$ $$g(x) = \frac{12}{1 - x}$$ We need to find the composition $gf(x)$, which means $g(f(x))$. 2. **Recall the composition formula:** $$gf(x) = g(f(x))$$ This means we substitute $f(x)$ into $g(x)$ wherever we see $x$. 3. **Substitute $f(x)$ into $g(x)$:** $$gf(x) = g(2x + 3) = \frac{12}{1 - (2x + 3)}$$ 4. **Simplify the denominator:** $$1 - (2x + 3) = 1 - 2x - 3 = -2x - 2$$ So, $$gf(x) = \frac{12}{-2x - 2}$$ 5. **Factor the denominator:** $$-2x - 2 = -2(x + 1)$$ 6. **Rewrite the function:** $$gf(x) = \frac{12}{-2(x + 1)}$$ 7. **Simplify the fraction by dividing numerator and denominator by 2:** $$gf(x) = \frac{\cancel{12}^{6}}{\cancel{-2}^{-1}(x + 1)} = \frac{6}{-1(x + 1)} = -\frac{6}{x + 1}$$ **Final answer:** $$gf(x) = -\frac{6}{x + 1}$$ This is the composition of $g$ and $f$ evaluated at $x$. Note: The domain of $gf$ excludes values that make the denominator zero, so $x \neq -1$.